A small object of mass m starts from rest at the position shown and slide along the frictionless loop-the-loop track of radius R. what is the smallest value of y such that object will slide without losing contact with the track ? (1) R/2, (2) R/4, (3) R, (4) 2R, (5) zero

A small object of mass m starts from rest at the position shown and slide along the frictionless loop-the-loop track of radius R. what is the smallest value of y such that object will slide without losing contact with the track ? (1) R/2, (2) R/4, (3) R, (4) 2R, (5) zero

PHSX 220 Homework 12 D2L – Due Thursday April 13 – 5:00 pm Exam 3 MC Review Problem 1: A 1.0-kg with a velocity of 2.0m/s perpendicular towards a wall rebounds from the wall at 1.5m/s perpendicularlly away from the wall. The change in the momentum of the ball is: A. zero B. 0.5 N s away from wall C. 0.5 N s toward wall D. 3.5 N s away from wall E. 3.5 N s toward wall Problem 2: A 64 kg man stands on a frictionless surface with a 0.10 kg stone at his feet. Both the man and the person are initially at rest. He kicks the stone with his foot so that his end velocity is 0.0017m/s in the forward direction. The velocity of the stone is now: A. 1.1m/s forward B. 1.1m/s backward C. 0.0017m/s forward D. 0.0017m/s backward E. none of these Problem 3: A 2-kg cart, traveling on a rctionless surface with a speed of 3m/s, collides with a stationary 4-kg cart. The carts then stick together. Calculate the magnitude of the impulse exerted by one cart on the other: A. 0 B. 4N s C. 6N s D. 9N s E. 12N s Problem 4: A disc has an initial angular velocity of 18 radians per second. It has a constant angular acceleration of 2.0 radians per second every second and is slowing at rst. How much time elapses before its angular velocity is 18 rad/s in the direction opposite to its initial angular velocity? A. 3.0 s B. 6.0 s C. 9.0 s D. 18 s E. 36 s Problem 5: Three point masses of M, 2M, and 3M, are fastened to a massless rod of length L as shown. The rotational inertia about the rotational axis shown is: A. (ML2=2) B. (ML2) C. (3ML2)=2 D. (6ML2) E. (3ML2)=4 Problem 6: A board is allowed to pivot about its center. A 5-N force is applied 2m from the pivot and another 5-N force is applied 4m from the pivot. These forces are applied at the angles shown in the gure. The magnitude of the net torque about the pivot is: A. 0 Nm B. 5 Nm C. 8.7 Nm D. 15 Nm E. 26 Nm Problem 7: A solid disk (r=0.03 m) and a rotational inertia of 4:5×10􀀀3kgm2 hangs from the ceiling. A string passes over it with a 2.0-kg block and a 4.0-kg block hanging on either end of the string and does not slip as the system starts to move. When the speed of the 4 kg block is 2.0m/s the kinetic energy of the pulley is: A. 0.15 J B. 0.30 J C. 1.0J D. 10 J E. 20 J Problem 8: A merry go round (r= 3.0m, I =600 kgm2) is initially spinning with an angular velocity of 0.80 radians per second when a 20 kg point mass moves from the center to the rim. Calculate the nal angular velocity of the system: A. 0.62 rad/s B. 0.73 rad/s C. 0.80 rad/s D. 0.89 rad/s E. 1.1 rad/s

PHSX 220 Homework 12 D2L – Due Thursday April 13 – 5:00 pm Exam 3 MC Review Problem 1: A 1.0-kg with a velocity of 2.0m/s perpendicular towards a wall rebounds from the wall at 1.5m/s perpendicularlly away from the wall. The change in the momentum of the ball is: A. zero B. 0.5 N s away from wall C. 0.5 N s toward wall D. 3.5 N s away from wall E. 3.5 N s toward wall Problem 2: A 64 kg man stands on a frictionless surface with a 0.10 kg stone at his feet. Both the man and the person are initially at rest. He kicks the stone with his foot so that his end velocity is 0.0017m/s in the forward direction. The velocity of the stone is now: A. 1.1m/s forward B. 1.1m/s backward C. 0.0017m/s forward D. 0.0017m/s backward E. none of these Problem 3: A 2-kg cart, traveling on a rctionless surface with a speed of 3m/s, collides with a stationary 4-kg cart. The carts then stick together. Calculate the magnitude of the impulse exerted by one cart on the other: A. 0 B. 4N s C. 6N s D. 9N s E. 12N s Problem 4: A disc has an initial angular velocity of 18 radians per second. It has a constant angular acceleration of 2.0 radians per second every second and is slowing at rst. How much time elapses before its angular velocity is 18 rad/s in the direction opposite to its initial angular velocity? A. 3.0 s B. 6.0 s C. 9.0 s D. 18 s E. 36 s Problem 5: Three point masses of M, 2M, and 3M, are fastened to a massless rod of length L as shown. The rotational inertia about the rotational axis shown is: A. (ML2=2) B. (ML2) C. (3ML2)=2 D. (6ML2) E. (3ML2)=4 Problem 6: A board is allowed to pivot about its center. A 5-N force is applied 2m from the pivot and another 5-N force is applied 4m from the pivot. These forces are applied at the angles shown in the gure. The magnitude of the net torque about the pivot is: A. 0 Nm B. 5 Nm C. 8.7 Nm D. 15 Nm E. 26 Nm Problem 7: A solid disk (r=0.03 m) and a rotational inertia of 4:5×10􀀀3kgm2 hangs from the ceiling. A string passes over it with a 2.0-kg block and a 4.0-kg block hanging on either end of the string and does not slip as the system starts to move. When the speed of the 4 kg block is 2.0m/s the kinetic energy of the pulley is: A. 0.15 J B. 0.30 J C. 1.0J D. 10 J E. 20 J Problem 8: A merry go round (r= 3.0m, I =600 kgm2) is initially spinning with an angular velocity of 0.80 radians per second when a 20 kg point mass moves from the center to the rim. Calculate the nal angular velocity of the system: A. 0.62 rad/s B. 0.73 rad/s C. 0.80 rad/s D. 0.89 rad/s E. 1.1 rad/s

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The Rocket Equation The Tsiolovsky Rocket Equation describes the velocity that results from pushing matter (exploding rocket fuel) in the opposite direction to the direction you want to travel. This assignment requires you to do basic calculation using the Tsiolovsky Rocket Equation : v[t] = eV Log M M – bR t  – g t The parameters used are : ◼ eV exhaust velocity (m/s) ◼ pL payload (kg) ◼ fL fuel load (kg) ◼ M is the mass of the rocket (pL+fL, kg) ◼ bR the burn rate of fuel (kg/s) ◼ g the force due to gravity ms2 The variables calculated are : h(t) the height of the rocket at time t (m) v(t) the velocity of the rocket at time t (m/s) m(t) the mass of the rocket at time t (kg) Questions Question 1 (1 mark) Write an expression corresponding to the Tsiolovsky rocket equation and use integrate to find a function to describe the height of the rocket during fuel burn. Question 2 (2 marks) The fuel burns at a constant rate. Find the time (t0), velocity (vmax), and height (h0) of the rocket when the fuel runs out (calculate the time when the fuel runs out, and substitute this into the height Printed by Wolfram Mathematica Student Edition and velocity equations). Question 3 (2 marks) The second phase is when the only accelaration acting on the rocket is from gravity. This phase starts from the height and velocity of the previous question, and the velocity is given by the projectile motion equation, v(t) = vmax – g (t – t0). Use Solve to find the time when this equation equals 0. This will be the highest point the rocket reaches before returning to earth. Question 4 (1 marks) Integerate the projectile motion equation and add h0 to find the maximum height the rocket reaches. Question 5 (1 marks) Use Solve over the projectile motion equation to find the time when the height is 0. 2 assignment4.nb Printed by Wolfram Mathematica Student Edition

The Rocket Equation The Tsiolovsky Rocket Equation describes the velocity that results from pushing matter (exploding rocket fuel) in the opposite direction to the direction you want to travel. This assignment requires you to do basic calculation using the Tsiolovsky Rocket Equation : v[t] = eV Log M M – bR t  – g t The parameters used are : ◼ eV exhaust velocity (m/s) ◼ pL payload (kg) ◼ fL fuel load (kg) ◼ M is the mass of the rocket (pL+fL, kg) ◼ bR the burn rate of fuel (kg/s) ◼ g the force due to gravity ms2 The variables calculated are : h(t) the height of the rocket at time t (m) v(t) the velocity of the rocket at time t (m/s) m(t) the mass of the rocket at time t (kg) Questions Question 1 (1 mark) Write an expression corresponding to the Tsiolovsky rocket equation and use integrate to find a function to describe the height of the rocket during fuel burn. Question 2 (2 marks) The fuel burns at a constant rate. Find the time (t0), velocity (vmax), and height (h0) of the rocket when the fuel runs out (calculate the time when the fuel runs out, and substitute this into the height Printed by Wolfram Mathematica Student Edition and velocity equations). Question 3 (2 marks) The second phase is when the only accelaration acting on the rocket is from gravity. This phase starts from the height and velocity of the previous question, and the velocity is given by the projectile motion equation, v(t) = vmax – g (t – t0). Use Solve to find the time when this equation equals 0. This will be the highest point the rocket reaches before returning to earth. Question 4 (1 marks) Integerate the projectile motion equation and add h0 to find the maximum height the rocket reaches. Question 5 (1 marks) Use Solve over the projectile motion equation to find the time when the height is 0. 2 assignment4.nb Printed by Wolfram Mathematica Student Edition

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COMP 4440/5440 – Dr. Erdemir Mobile Robotics Project (DUE 12/02/2015) HONOR CODE I pledge my honor that I have neither given nor received aid on this work. Do not sign until after you have completed your assignment. Name: Signature: 1. (Prerequisite) Given the asset package (it is in mytsu under assignments folder) download it, open a new project (don’t double click on the file, it won’t open), go to Assets/Import Package/Custom Package, select the asset package given to you (project.unitypackage). After the import is completed, you will see main scene in the assets folder of your project. Double click on it, and choose “Don’t Save” option if it asks for save. 2. Print this page and attach it to your code and your snapshot of your final scene (5 points) 3. After the class starts, instructor will come next to you and you are supposed to show him your code running (10 points) 4. When you run the code you will see, your robot (red cube that we used in the class) is trying to reach the targets but it can’t due to an obstacle between your robot and the targets. Write a code that makes this robot to avoid from the obstacles and reach the targets. Your code will read the collision and intelligently avoid from other moving robots and fixed obstacles and get the three targets. (50 points) 5. Maximum time is 2 minutes, your robot shall get the targets in 2 minutes (10 points) 6. Your robot shall escape from the blue robot and not collide them. (10 Points) 7. Anything extra (up to 20 points) ? Moving objects, new sensor, Artificial Intelligence or other techniques. 8. YOU CAN NOT USE TRANSLATE FUNCTION. USE ONLY AddRelativeForce FUNCTION IN THE FORWARD DIRECTION AS ALL THE MOBILE ROBOTS WORK.

COMP 4440/5440 – Dr. Erdemir Mobile Robotics Project (DUE 12/02/2015) HONOR CODE I pledge my honor that I have neither given nor received aid on this work. Do not sign until after you have completed your assignment. Name: Signature: 1. (Prerequisite) Given the asset package (it is in mytsu under assignments folder) download it, open a new project (don’t double click on the file, it won’t open), go to Assets/Import Package/Custom Package, select the asset package given to you (project.unitypackage). After the import is completed, you will see main scene in the assets folder of your project. Double click on it, and choose “Don’t Save” option if it asks for save. 2. Print this page and attach it to your code and your snapshot of your final scene (5 points) 3. After the class starts, instructor will come next to you and you are supposed to show him your code running (10 points) 4. When you run the code you will see, your robot (red cube that we used in the class) is trying to reach the targets but it can’t due to an obstacle between your robot and the targets. Write a code that makes this robot to avoid from the obstacles and reach the targets. Your code will read the collision and intelligently avoid from other moving robots and fixed obstacles and get the three targets. (50 points) 5. Maximum time is 2 minutes, your robot shall get the targets in 2 minutes (10 points) 6. Your robot shall escape from the blue robot and not collide them. (10 Points) 7. Anything extra (up to 20 points) ? Moving objects, new sensor, Artificial Intelligence or other techniques. 8. YOU CAN NOT USE TRANSLATE FUNCTION. USE ONLY AddRelativeForce FUNCTION IN THE FORWARD DIRECTION AS ALL THE MOBILE ROBOTS WORK.

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

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Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

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